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The Quadratic Reciprocity Theorem

We are now going to observe a relationship between the Legendre symbols$(p/q)$ and $(q/p)$ where p and q are odd primes. First, let's start looking at the relationship between the Legendre symbols of two odd primes with the following table showing the Legendre symbol $(p/q)$ from p = 19 to q = 17.

$(p/q)$

p = 5

p = 7

p = 11

p = 13

p = 17

p = 19

q = 3

- 1

1

-1

1

-1

1

q = 5

-

-1

1

-1

-1

1

q = 7

-

-

1

-1

-1

-1

q = 11

-

-

-

-1

-1

-1

q = 13

-

-

-

-

1

-1

q = 17

-

-

-

-

-

1

And now lets look at the next table showing the Legendre symbol $(q/p)$ from p = 19 to q = 17. The additional highlighted entries are different from $(p/q)$:

$(q/p)$

q = 5

q = 7

q = 11

q = 13

q = 17

q = 19

p = 3

- 1

-1

1

1

-1

-1

p = 5

-

-1

1

-1

-1

1

p = 7

-

-

-1

-1

-1

1

p = 11

-

-

-

-1

-1

1

p = 13

-

-

-

-

1

-1

p = 17

-

-

-

-

-

1

From these tables, we deduce the following relationships from these Legendre symbols:

(5/3) = (3/3)

-

-

-

-

-

(7/3) = -(3/7)

(7/5) = (5/7)

-

-

-

-

(11/3) = -(3/11)

(11/5) = (5/11)

(11/7) = - (7/11)

-

-

-

(13/3) = (3/13)

(13/5) = (5/13)

(13/7) = (7/13)

(13/11) = (11/13)

-

-

(17/3) = (3/17)

(17/5) = (5/17)

(17/7) = (7/17)

(17/11) = (11/17)

(17/13) = (13/17)

-

(19/3) = -(3/19)

(19/5) = (5/19)

(19/7) = -(7/19)

(19/11) = -(11/19)

(19/13) = (13/19)

(19/17) = (17/19)

We can deduce $(p/q) = -(q/p)$ if $p \equiv q \equiv 3 \pmod 4$. If just one of p or q is congruence to 1 (mod 4), then we can deduce that $(p/q) = (q/p)$. We will summarize this in the Quadratic Reciprocity theorem.

Now we know that $(q - 1)^{\frac{p^2 - 1}{8}}$ is even, and that $2S$ is even. Hence $p(S(p, q) - g)$ must also be even. Since p is an odd prime, then $(S(p, q) - g )$ must be even, so it follows that either $S(p, q)$ and $-g$ are both odd or both even. Hence: